# Proof of the equivalence between two ways of defining ESS

## Background

Two common ways of defining what an Evolutionary Stable Strategy (ESS) are:

First definition: Consider a population composed of populations playing two strategies, $\mathbf{p}$ and $\mathbf{q}$. Let us denote $W(\mathbf{p})$ the average fitness of the strategy $\mathbf{p}$. A population consisting of individuals playing $\mathbf{p}$ will be an ESS if, whenever a small amount of deviant individuals playing $\mathbf{q}$, the old type $\mathbf{p}$ fares batter than the newcomers $\mathbf{q}$. This means that for all $\mathbf{p}\neq \mathbf{q}$, \begin{equation} W(\mathbf{p}) > W (\mathbf{q}) \end{equation}

Second definition: $E(\mathbf{p},\mathbf{q})$ is the payoff for $\mathbf{p}$-strategist against a $\mathbf{q}$-strategist. The strategy $\mathbf{p}$ is an ESS if and only if the following conditions are satisfied:

1. $E(\mathbf{p},\mathbf{p})\geq E(\mathbf{q},\mathbf{p}) \quad \forall \mathbf{q}$

2. If $\mathbf{q}\neq \mathbf{p}$ and $E(\mathbf{p},\mathbf{p}) = E(\mathbf{q},\mathbf{p})$, then $E(\mathbf{p},\mathbf{q})> E(\mathbf{q},\mathbf{q})$

## Question (shorter version)

The above two ways of defining what an ESS should be equivalent. How can I prove such an equivalence? Alternatively, do you know of a book (or a paper) that has this proof?

## Question (longer version)

I understand that the first definition implies the second. According to the first definition: \begin{align*} W(\mathbf{p}) - W (\mathbf{q}) &>0\\ (1-\epsilon) [E(\mathbf{p},\mathbf{p}) - E(\mathbf{q},\mathbf{p})] + \epsilon [E(\mathbf{p},\mathbf{q}) - E(\mathbf{q},\mathbf{q})] &>0 \end{align*} where $\epsilon$ is frequency of individuals playing $\mathbf{q}$. Since $0<\epsilon<1$, if the above inequality is true, the conditions 1 and 2 in the second definition must be true. The part that I find less clear is how the second definition implies the first definition. More specifically, that if $\mathbf{p}$ is a strict Nash equilibrium (i.e., $E(\mathbf{p},\mathbf{p})> E(\mathbf{q},\mathbf{p})$), then $W(\mathbf{p}) > W (\mathbf{q})$.

• Aren't these definitions applicable to different models of evolution? It seems like the first is applicable in a case where each strategy has, regardless of the other strategies, a known distribution of payoffs. The second seems to have a payoff which is dependent not only on the strategy but also on the other players' strategy. If they are different models, I am not sure what it means to be equivalent. – Hans Dec 12 '15 at 4:20
• @Hans The equivalence between these defs is mentioned in a few textbooks, such as p.232 of Gintis Game Theory Evolving and on p. 63 of Hofbauer & Sigmund's Evolutionary Games and Population Dynamics. – falsum Dec 12 '15 at 4:40

We need to show that if $\mathbf{p}$ is a strict Nash equilibrium, $E(\mathbf{p},\mathbf{p})> E(\mathbf{q},\mathbf{p})$, then $W(\mathbf{p}) > W (\mathbf{q})$ for some sufficiently small $\epsilon$. When $E(\mathbf{p},\mathbf{q}) \geq E(\mathbf{q},\mathbf{q})$, $\mathbf{p}$ clearly dominates $\mathbf{q}$.The less obvious case is when $E(\mathbf{p},\mathbf{q}) < E(\mathbf{q},\mathbf{q})$.
Since $E(\mathbf{p},\mathbf{p}) - E(\mathbf{q},\mathbf{p})>0$, there should be a strictly positive number $k$ such that $$k[E(\mathbf{p},\mathbf{p}) - E(\mathbf{q},\mathbf{p})]+ E(\mathbf{p},\mathbf{q}) - E(\mathbf{q},\mathbf{q})>0$$ is true. Whatever this number is, we can write $k$ as $(1-\epsilon)/\epsilon$ for $0<\epsilon<1$. Substituting this into the equation above gives us the desired result.